Question

I need to answer #3

could be done in only one way, we see that if we take the table for G and rename the identity e, the next element listed a, and the last element b, the resulting table for G must be the same as the one we had for G. As explained in Section 3, this renaming gives an isomorphism of the group G' with the group G. Definition 3.7 defined the notion of isomorphism and of isomorphic binary structures. Groups are just certain types of binary structures, so the same definition pertains to them. Thus our work above can be summarized by saying that all groups with a single element are isomorphic, all groups with just two clements are isomorphic, and all groups with just three elements are isomorphic. We use the phrase up to isomorphism to express this identification using the equivalence relation . Thus we may say. "There is only one group of three elements, up to isomorphism 4.19 Table 4.20 Table 4.21 Table EXERCISES 4 Computations In Exercises 1 through 6, determine whether the binary operation gives a group structure on the given set. If no group results, give the first axiom in the order 2 from Definition 4.1 that does not hold. 14 Let * be defined on Z by letting a * b = ab. 2) Let * be defined on 22-Zn 1 n E Z} by letting a * b = a + b. 3. Let s be defined on Rt by letting a *b ah. watcn accoc. 4, Let * be defined on Q by letting a * b = ab. 5. Let * be defined on the set R* of nonzero real numbers by letting ab- ajb 6. Let * be defined on C by letting a b labl. 7 Give an example of an abelian group G where G has exactly 1000 elements. 8. We can also consider multiplication ,n modulo n in Zn. For example, 5 7 6 = 2 in Z7 because 5-6-30 4(7) +2. The set {1, 3, 5, 73 with multiplication 's modulo 8 is a group. Give the table for this group 9. Show that the group (U, ) is not isomorphic to either (IR, ) or (R.). (All three groups have cardinality [RI) 10. Let n be a positive integer and let nZ = {nm 1 m e Z}. a. Show that (nZ, + is a group.l Ko a b. Show that (nZ, (Z, +)

Answer 1

Let us check group axioms as follow i) closureness: - a times b elementof R^+ forall a, b elementof R^+ let a, b elementof R^* then a > 0^b > 0 so ab . 0 & squareroot ab elementof R^+ therefore a * b = squareroot ab elementof R^+ hence it is closed ii) associative: - a times (b times c) = (a times b) times c Forall a, b, c elementof R^+ let a, b. c elementof R% + then a times (b times c) = a times (squareroot bc) = a squareroot bc rightarrow 1 (a times b) times c = ab times c = squareroot squareroot ab middot c righta ii since 1 & 2 are not equal squareroot a squareroot bc notequalto squareroot squareroot ab middot c therefore a times (b times c) notequalto (a times b) times c hence not associate therefore R^+ is not group with binary operation